Question

A train starts from rest at station A and accelerates at 0.5m/s^2 for 60 seconds. Afterwards it travels with a constant velocity for 15 minutes. It then decelerates at 1m/s^2 until t is brought to rest at station B. Determine the distance between the stations

Answer 1

Answer: Total distance = 28350 m

Explanation: We can divide the distance travel in 3 steps: one accelerating, the middle step at constant velocity and a final one decelerating.

For the first distance (x1) , we know the acceleration from rest (0.5 m/sec²) and the time traveled (60 sec).

By definition, acceleration is equal to the change in velocity over time, so  we can put the following:

a= vf-vo/Δt⇒ if vo=0 (starts from rest) ⇒vf= a*Δt=0.5 m/sec²*60 sec=30 m/sec.

Starting from rest, we can write that x1= 1/2 * a* Δt²= 900 m.

(we arrive to the same result applying vf²=2*a*Δx).

For the second part, we know that at t=60 sec, v= 30 m/sec.

As the second part is traveled at constant velocity, by definition, we can write the following:

v= Δx/Δt  (Δt=15 min= 15*60 sec/min=900 sec) ⇒Δx=v*Δt=30m/sec*900sec=27000 m

FInally, as the train is brought to rest after decelerating, and we know that it's decelerating at a rate of 1 m/sec2, and the initial velocity is 30 m/sec, we can get the time traveled while it was decelerating:

vf-vo= a*Δt   ⇒  Δt=(vf-vo)/a ⇒Δt= 30 sec 9 (as vf=0)

With this information, we can use the following kinematic equation:

Δx= vo*t + 1/2 * a* t²

Replacing by the values that we have got previously, we got :

Δx= 900 m - 450 m = 450 m

Adding the three distances, we have:

Δx= 900 m + 27000 m + 450 m = 28350 m.

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